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Tanzania [10]
2 years ago
15

A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of

dimensions 5.00 cm by 5.22 cm that rotates in an adjustable magnetic field. What is the field strength needed to produce a 24.0 V peak emf
Physics
1 answer:
Sav [38]2 years ago
6 0

The field strength needed to produce a 24.0 V peak emf is 0.73T.

To find the answer, we need to know about the expression of emf.

What's the expression of peak emf produced in a rotating rectangular loops?

  • The peak emf produced in a rotating loops= N×B×A×w
  • N= no. of turns of the loop, B= magnetic field, A= area of loop and w= angular frequency
  • So, B = emf/(N×A×w)
<h3>What's the magnetic field applied to the loop, when rectangular coil with 300 turns of dimensions 5.00 cm by 5.22 cm rotates at 400 rpm produce a 24.0 V peak emf?</h3>
  • N= 300, A= 5cm × 5.22cm = 0.05m × 0.0522m = 0.00261 m²
  • Emf= 24V, w= 2π×400 rpm= 2π×(400rps/60) = 42 rad/s
  • Now, B= 24/(300×0.00261×42)

B= 24/(300×0.00261×42) = 0.73T

Thus, we can conclude that the magnetic field is 0.73T.

Learn more about the electromagnetic force here:

brainly.com/question/13745767

#SPJ4

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Answer with Explanation:

We are given that

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Torque=r\times F

Substitute the values then we get

Torque= (0.5j-2k)\times (2i-3k)

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By using i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0

b.r=2i-3k

r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k

Torque about point (2,0,-3)

\tau=(-2i+0.5j+k)\times (2i-3k)

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where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.

In this problem, the first medium is the glass (n_1 = 1.50), while the second medium is oil (n_2 =1.46), therefore the critical angle is given by
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n a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section of rad
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Answer: 1477.78 N

Explanation:

Let's assume that the cross sectional area of the smaller piston be A1

let's also assume the cross sectional area of the larger piston be A2

We assume the force applied to the smaller piston be F1

We also assume the force applied to the larger piston be F2

we then use the formula

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From our question,

The radius of the smaller piston is 5 cm = 0.05 m

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The force of the larger piston is 13300 N

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F1 = (13300 * 0.007855) / 0.070695

F1 = 104.4715 / 0.070695

F1 = 1477.78 N

Thus, the force the compressed air must exert is 1477.78 N

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